Re: integrate (1+2e^x-e^(-x)^(-1)

Re: integrate (1+2e^x-e^(-x)^(-1)

In the denominator I have 2u^2+u-1. The trigonometric substitution requires to have just two elements in the polynom - a squared variable and a squared constant.Here I have a third member - u, where do you propose it should go?

Re: integrate (1+2e^x-e^(-x)^(-1)

You could complete the square in the denominator.

Here lies the reader who will never open this book. He is forever dead.Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and PunishmentThe knowledge of some things as a function of age is a delta function.

Re: integrate (1+2e^x-e^(-x)^(-1)

Re: integrate (1+2e^x-e^(-x)^(-1)

You get it to the form (a*x+b)^2+c.

Here lies the reader who will never open this book. He is forever dead.Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and PunishmentThe knowledge of some things as a function of age is a delta function.

Re: integrate (1+2e^x-e^(-x)^(-1)

mmmm.....2u^2+u-1 does not have an (a*x+b)^2+c form or I cannot find it.2u^2+u-1 = (2u-1)(u+1), and this I used in a "partial fractions" approach. But I cannot find any square form for this particular polynomial.

Re: integrate (1+2e^x-e^(-x)^(-1)

Of course, I am just showing that it can also be done as zf. suggested. The partial fractions work better in this integral.

Last edited by anonimnystefy (2013-02-18 11:43:46)

Here lies the reader who will never open this book. He is forever dead.Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and PunishmentThe knowledge of some things as a function of age is a delta function.